Difference between revisions of "2024 SSMO Speed Round Problems/Problem 6"
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+ | Let <math>a_n</math> equal <math>1</math> if the nth house is occupied and <math>0</math> if the nth house is unoccupied. Thus, we are trying to find <math>\mathbb{E}[a_1+a_2+a_3+a_4] = 4\mathbb{E}[a_1]</math> by the Linearity of Expectation. Instead of calculating the probability that a single house is occupied, we instead calculate the probability that it is unoccupied. Each person has a <math>\frac{1}{4}</math> chance of residing in that house, and thus a <math>\frac{3}{4}</math> chance of not residing in that house. The probability that none of the people stay in that house is <math>\left (\frac{3}{4} \right)^4 = \frac{81}{256}</math>. Thus, the probability that any given house is occupied, or <math>\mathbb{E}[a_1]</math>, is <math>1 - \frac{81}{256} = \frac{175}{256}</math>. Thus, our answer is <math>4\cdot \frac{175}{256} = \frac{175}{64}</math>, <math>175+64=\boxed{239}</math>. | ||
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+ | -Vivdax |
Latest revision as of 19:41, 2 May 2025
Problem
There are people and
houses. Each person independently randomly chooses a house to live in. The expected number of inhabited houses can be expressed as
, where
and
are relatively prime positive integers. Find
.
Solution
Let equal
if the nth house is occupied and
if the nth house is unoccupied. Thus, we are trying to find
by the Linearity of Expectation. Instead of calculating the probability that a single house is occupied, we instead calculate the probability that it is unoccupied. Each person has a
chance of residing in that house, and thus a
chance of not residing in that house. The probability that none of the people stay in that house is
. Thus, the probability that any given house is occupied, or
, is
. Thus, our answer is
,
.
-Vivdax